Correct Answer - C
`x = sqrt(2+sqrt(2-sqrt(2+x)))`
We must have `x=2-sqrt(2+ x)ge 0`
`rArr 2 ge sqrt(2+x)`
`rArr x le 2`
Also `x le 2`
`rArr x ge -2`
`therefore x in [-2,2]`
Let `x = 2 cos theta`, where `theta in [0, pi]`. Then,
`x = wsqrt(2+sqrt(2-sqrt(2+2cos theta)))`
`rArr 2cos theta = sqrt(2+sqrt(2-2cos.(theta)/(2)))`
`= sqrt(2+sqrt(2(1-cos.(theta)/(2))))`
`= sqrt(2+ 2 sin.(theta)/(4))`
`sqrt(2+2cos((pi)/(2)-(theta)/(4)))`
`= sqrt(2(1+cos((pi)/(2)-(theta)/(4))))`
`rArr 2 cos theta = 2 cos ((pi)/(4)-(theta)/(8))`
`rArr theta = (pi)/(4)-(theta)/(8)`
`rArr theta = (2pi)/(9)`
Hence, `x=2 cos.(2pi)/(9)=2 cos 40^(@)`
